0070._Climbing_Stairs.md

70. Climbing Stairs

难度: Easy

刷题内容

原题连接

• https://leetcode.com/problems/climbing-stairs/description/

内容描述

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Example 2:

Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

解题方案

思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******

Fibonacci 的DP版本

Top down， 记忆化搜索, beats 96.64%

这里cache用dict，用array也一样。

class Solution(object):
    cache = {
        1: 1,
        2: 2
    }
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n in self.cache:
            return self.cache[n]
        self.cache[n] = self.climbStairs(n-1) + self.climbStairs(n-2)
        return self.cache[n]

思路 2 - 时间复杂度: O(N)- 空间复杂度: O(N)******

采用bottom up思想, 动态规划，beats 96.64%

class Solution(object):
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        cache = [1] * (n+1)
        for i in range(2, n+1):
            cache[i] = cache[i-1] + cache[i-2]
        return cache[-1]

思路 3 - 时间复杂度: O(N)- 空间复杂度: O(1)******

当然用bottom up还有一点，可以只存每次最后两个数，可以save space.，这样就只用到constant space.

永远只存3个元素，save space, beats 96.64%

class Solution(object):
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        cache = [1, 2, 3]
        if n < 4:
            return cache[n-1]
        for i in range(3, n+1):
            cache[2] = cache[0] + cache[1]
            cache[0], cache[1] = cache[1], cache[2]
        return cache[2]

思路 4 - 时间复杂度: O(lgN)- 空间复杂度: O(1)******

另外还有一个公式法：

由于这里面相当于standard Fibonacci函数向前进了一步，排列为1，2，3，5而非原本的1，1，2，3，所以代码中使用n+1

时间复杂度为O(lgN)是因为pow(X, Y)函数的时间复杂度:

• 当 Y < 2^63, lgY
• 当 Y >= 2^63, e^(Y * lgX)

class Solution(object):
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        sqrt5 = math.sqrt(5)
        fibn = pow((1 + sqrt5) / 2, n+1) - pow((1 - sqrt5) / 2, n+1)
        return int(fibn/sqrt5)